I am having trouble understanding the derivation of the rate of spontaneous and stimulated emission given in this link.

We have a perturbation that takes the form: $$ \hat H=\sum_{\vec k}f(\vec r,\vec k) e^{-i\omega_k t}$$ With an initial and final been: $$\left|i\right>=\left|a\right> \left|...n_{\vec l}...\right>$$ $$\left|f\right>=\left|b\right> \left|...n_{\vec l}+1...\right>$$ Where the difference in energy between the initial and final states is given by: $$E_f-E_i=E_b-E_a +\hbar \omega_l$$

Now I know that from Fermi's golden rule, for a potential of the form $\hat V=V_0 e^{-i\omega t}$ we have a transition probability of the form: $$P_{if}=\frac{2\pi t |V_{fi}|^2}{\hbar^2} \delta(\hbar \omega-E_{fi})$$ So in our case I would expect us to have $$P_{if}=\frac{2\pi t }{\hbar^2} \sum_{\vec k} |f_{fi}|^2\delta(\hbar \omega_k-(E_b-E_a +\hbar \omega_l)) \tag{1} \label{1}$$ However in the link they appear to be using: $$P_{if}=\frac{2\pi t }{\hbar^2} \sum_{\vec k} |f_{fi}|^2\delta(\hbar \omega_k+E_b-E_a) \tag{2} \label{2}$$ Intuitively (\ref{2}) seems right to me, but this seems inconsistent with (\ref{1}), which I cannot see why it is wrong. So, if can (\ref{2}) be derived from (\ref{1}), if so how and if not what why is (\ref{1}) wrong?


I think the answer to this question may lie in the 'picture' of quantum mechanics used. In the Schrödinger picture the interaction Hamiltonian is time independent. Now if for the form of Fermi's golden rule I have given here $\hat V$ must be written in the Schrödinger picture then equation (\ref{2}) does follow naturally from it. If however it is written in the interaction picture then (\ref{1}) follows naturally from it. Thus am I correct in saying that for Fermi's golden rule in the form I have given $\hat V$ must be written in the Schrödinger picture?

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    $\begingroup$ The total energy in the initial and final states should be the same right? $\endgroup$ Sep 12, 2016 at 13:03
  • $\begingroup$ @flippiefanus Yes, which would indicate expression (1) is wrong, but mathematically why is it wrong in the context of Fermi's Golden Rule? $\endgroup$ Sep 12, 2016 at 14:01
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    $\begingroup$ The problem then already lies with the 2nd prior expression, (4th expression from the start) where the left had side should be zero. However, how this then eventually gives your final expression I don't know. I'll think about it. $\endgroup$ Sep 13, 2016 at 4:17
  • $\begingroup$ Compare to this. $\endgroup$ May 27, 2020 at 20:09
  • $\begingroup$ Related: physics.stackexchange.com/q/268694/226902 $\endgroup$
    – Quillo
    Oct 3, 2023 at 15:01

1 Answer 1


The problem is clearly that the energy of the photon enters twice, the question is why is that so. In Fermi's golden rule for a quantum system but external classical light field you have a perturbation operator $$ H_p = E_0(r) e^{-i \omega_l t} $$ Transitions occur when $\omega_l$ is resonant to a transition frequency of the system. The delta function becomes $$ \delta(\hbar\omega_l-E_{fi}) $$ or in words $$ \text{external frequency} = \text{transition frequency} $$ Clearly energy conservation is violated because $E_{fi}\neq 0$, which is ok because the hamiltonian of the system is time dependent. In other words the energy in the electromagnetic field is not included in our total energy and so neither is conserved.

What happens for a quantized electromagnetic field? For a quantized field the hamiltonian should not be time dependent because all the time evolution is included in the system, especially the time evolution of the electromagnetic field. For a stationary hamiltonian the delta function in Fermi's golden rule becomes $\delta(E_{fi})$, which implies conservation of energy.

The error in your derivation is, as you have noted yourself, that somewhere Schroedinger and Heisenberg picture were incorrectly mixed.

  • $\begingroup$ Sakurai's classic QM text contrasts the two pictures and dramatizes the contrast between constant and periodic perturbations in the formula. $\endgroup$ May 27, 2020 at 21:15

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